/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Express each repeating decimal a... [FREE SOLUTION] | 91Ó°ÊÓ

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Express each repeating decimal as a quotient of integers. If possible, reduce to lowest terms. \(0 . \overline{81}\)

Short Answer

Expert verified
\(0 . \overline{81} = \frac{9}{11}\)

Step by step solution

01

Setting up the equation

Let's denote the repeating decimal as a variable, \(x\). So, we can say \(x = 0 . \overline{81}\). Given that the decimal repeats every two digits, we multiply both sides of the equation by 100, getting \(100x = 81 . \overline{81}\).
02

Creating equation without decimal repetition

By subtracting the original equation from the one we obtained in step 1 (so, \(100x - x = 81 . \overline{81} - 0 . \overline{81}\) ), we can create an equation that does not have decimal repetition. This simplifies to \(99x = 81\).
03

Solve for x

By solving the equation \(99x = 81\) for \(x\), we get \(x = \frac{81}{99}\).
04

Reducing to the lowest term

Our final step would be to reduce \(x = \frac{81}{99}\) to its lowest term which can be done by identifying a common factor that divides both the numerator and the denominator. Upon doing so, we find that both \(81\) and \(99\) can be divided by \(9\) to give us the fraction \(\frac{9}{11}\).

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Most popular questions from this chapter

The sum, \(S_{n}\), of the first n terms of an arithmetic sequence is given by $$ S_{n}=\frac{n}{2}\left(a_{1}+a_{n}\right), $$ in which \(a_{1}\) is the first term and \(a_{n}\) is the nth term. The sum, \(S_{n}\), of the first \(n\) terms of a geometric sequence is given by $$ S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}, $$ in which \(a_{1}\) is the first term and \(r\) is the common ratio \((r \neq 1)\). Determine whether each sequence is arithmetic or geometric. Then use the appropriate formula to find \(S_{10}\), the sum of the first ten terms. \(7,19,31,43, \ldots\)

You will develop geometric sequences that model the population growth for California and Texas, the two most populated U.S. states. The table shows the population of California for 2000 and 2010 , with estimates given by the U.S. Census Bureau for 2001 through \(2009 .\) $$ \begin{array}{|l|l|l|l|l|l|l|} \hline \text { Year } & \mathbf{2 0 0 0} & \mathbf{2 0 0 1} & \mathbf{2 0 0 2} & \mathbf{2 0 0 3} & \mathbf{2 0 0 4} & \mathbf{2 0 0 5} \\ \hline \begin{array}{l} \text { Population } \\ \text { in millions } \end{array} & 33.87 & 34.21 & 34.55 & 34.90 & 35.25 & 35.60 \\ \hline \end{array} $$ $$ \begin{array}{|l|l|l|l|l|l|} \hline \text { Year } & \mathbf{2 0 0 6} & \mathbf{2 0 0 7} & \mathbf{2 0 0 8} & \mathbf{2 0 0 9} & \mathbf{2 0 1 0} \\ \hline \begin{array}{l} \text { Population } \\ \text { in millions } \end{array} & 36.00 & 36.36 & 36.72 & 37.09 & 37.25 \\ \hline \end{array} $$ a. Divide the population for each year by the population in the preceding year. Round to two decimal places and show that California has a population increase that is approximately geometric. b. Write the general term of the geometric sequence modeling California's population, in millions, \(n\) years after \(1999 .\) c. Use your model from part (b) to project California's population, in millions, for the year 2020 . Round to two decimal places.

Find the indicated term for the geometric sequence with first term, \(a_{1}\), and common ratio, \(r\). Find \(a_{8}\), when \(a_{1}=6, r=\frac{1}{2}\).

Write the first six terms of the geometric sequence with the first term, \(a_{1}\), and common ratio, \(r\). \(a_{1}=\frac{1}{4}, r=2\)

Find the indicated term for the geometric sequence with first term, \(a_{1}\), and common ratio, \(r\). Find \(a_{6}\), when \(a_{1}=18, r=-\frac{1}{3}\).

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